Simplify and expand the following expression: $ \dfrac{2}{3r + 30}- \dfrac{5}{3r + 6}+ \dfrac{4r}{r^2 + 12r + 20} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3r + 30} = \dfrac{2}{3(r + 10)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3r + 6} = \dfrac{5}{3(r + 2)}$ We can factor the quadratic in the third term: $ \dfrac{4r}{r^2 + 12r + 20} = \dfrac{4r}{(r + 10)(r + 2)}$ Now we have: $ \dfrac{2}{3(r + 10)}- \dfrac{5}{3(r + 2)}+ \dfrac{4r}{(r + 10)(r + 2)} $ The least common multiple of the denominators is: $ 9(r + 10)(r + 2)$ In order to get the first term over $9(r + 10)(r + 2)$ , multiply by $\dfrac{3(r + 2)}{3(r + 2)}$ $ \dfrac{2}{3(r + 10)} \times \dfrac{3(r + 2)}{3(r + 2)} = \dfrac{6(r + 2)}{9(r + 10)(r + 2)} $ In order to get the second term over $9(r + 10)(r + 2)$ , multiply by $\dfrac{3(r + 10)}{3(r + 10)}$ $ \dfrac{5}{3(r + 2)} \times \dfrac{3(r + 10)}{3(r + 10)} = \dfrac{15(r + 10)}{9(r + 10)(r + 2)} $ In order to get the third term over $9(r + 10)(r + 2)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{4r}{(r + 10)(r + 2)} \times \dfrac{9}{9} = \dfrac{36r}{9(r + 10)(r + 2)} $ Now we have: $ \dfrac{6(r + 2)}{9(r + 10)(r + 2)} - \dfrac{15(r + 10)}{9(r + 10)(r + 2)} + \dfrac{36r}{9(r + 10)(r + 2)} $ $ = \dfrac{ 6(r + 2) - 15(r + 10) + 36r} {9(r + 10)(r + 2)} $ Expand: $ = \dfrac{6r + 12 - 15r - 150 + 36r}{9r^2 + 108r + 180} $ $ = \dfrac{27r - 138}{9r^2 + 108r + 180}$ Simplify: $ = \dfrac{9r - 46}{3r^2 + 36r + 60}$